Thread: What did I did wrong in this U-sub Arc-Sec problem?

(cos 3x) / sin3x * (√sin²(3x) -2)


incase you aren't familiar with it, the formula for arc sec is

1/a arcsec u/a + c

a = -√2
u = sin 3x
du = 3 cos 3x

so 1/3 du = cos 3x dx

so I cancel out the top

and make since it's in the proper format I make it

I do 1/3 * a to get -√2/3 arcsec sin3x / -√2 + c

what did I do wrong, if anything?

I suggest that you use the u-sub. first; then, use the trig sub. second. Then resub back in terms of x. You are correct, you divide the 3 both side.

Originally Posted by Cbarker1

I suggest that you use the u-sub. first; then, use the trig sub. second. Then resub back in terms of x.

I appreciate the response but I'm very confused.

How did you derive sin(3x) to get 1/3 cos 3x?

sorry, maybe I'm being really stupid but I thought it'd be 3cos(3x)

Originally Posted by Cbarker1

I suggest that you use the u-sub. first; then, use the trig sub. second. Then resub back in terms of x.

Don't stress too much btw, I'll see the answer key tommorow but it was just bothering me I couldn't get it